Pandigital, twice

Three days ago Claudio Meller asked for solutions to a (base ten) pandigital number being expressed as the sum of two powers. Letting n = a^x + b^y, I assumed a, b, x, y > 1 and — since a^(m*x) = (a^m)^x — I did not bother to look for different ways of expressing the same power. Additionally, I restricted a and b to be less than 100. Even so, I came up with 15 solutions for the nine-digit 1-9 pandigitals and 16 solutions for the 0-9 pandigitals. Without any restriction on the size of a and b there would be far too many solutions to be of any interest. Which led me to wonder if there existed solutions where the concatenation of the digits of a, b, x, y was also pandigital.

For the 1-9 pandigitals, 3 solutions were readily computed:

291458637 = 581^3 + 9764^2
463985721 = 716^3 + 9845^2
513796482 = 761^3 + 8549^2

For the 0-9 pandigitals, 21 solutions (taking me the better part of two days):

2615307489 = 180^4 + 39567^2

3251746809 = 680^3 + 54197^2

3257904681 = 698^3 + 54017^2

3480912576 = 140^3 + 58976^2

3674092518 = 149^3 + 60587^2

3725804169 = 617^3 + 59084^2

4250687913 = 509^3 + 64178^2

4357028169 = 140^3 + 65987^2

5260389417 = 698^3 + 70145^2

5298340617 = 698^3 + 70415^2

5836290741 = 189^4 + 67530^2

6182053974 = 149^3 + 78605^2

6309257148 = 158^3 + 79406^2

6547231908 = 608^3 + 79514^2

6817924305 = 156^4 + 78903^2

7035892146 = 183^4 + 76905^2

8015623479 = 167^3 + 89504^2

9021853476 = 860^3 + 91574^2

9127068345 = 306^4 + 18957^2

9157023684 = 140^3 + 95678^2

9718352640 = 104^3 + 98576^2